$3$ particles $A, B$ and $C$ are projected from ground. According to the diagram, which of the following can be inferred about time of flight of $A (T A), B (T B)$ and $C (T C)$ ?

T_{A} = T_{B} < T_{C}

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T_{A} < T_{B} < T_{C}

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T_{A} > T_{B} > T_{C}

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T_{A} = T_{B} = T_{C}

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Solution

The correct option is D $T_{A} = T_{B} = T_{C}$
From the diagram we can infer, $H_{m a x}$ same for all the three particles.

$H_{m a x} = \frac{u^{2} {sin}^{2} θ}{2 g} \to \frac{u_{y}^{2}}{2 g}$
$T_{f} = \frac{2 u sin θ}{g} \to \frac{2 u_{y}}{g}$

As $H_{m a x}$ is same, we can say that $u_{y}$ is also same for all partciles.
$∴$ Time of flight is same for all particles.

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3 particles A,B and C are projected from ground. According to the diagram, which of the following can be inferred about time of flight of A(TA),B(TB) and C(TC)?

The correct option is D TA=TB=TCFrom the diagram we can infer, Hmax same for all the three particles. Hmax=u2sin2θ2g→u2y2g Tf=2usinθg→2uyg As Hmax is same, we can say that uy is also same for all partciles. ∴ Time of flight is same for all particles.

$3$ particles $A, B$ and $C$ are projected from ground. According to the diagram, which of the following can be inferred about time of flight of $A (T A), B (T B)$ and $C (T C)$ ?